Problem: $\dfrac{ -6p + 5q }{ -10 } = \dfrac{ 6p - 9r }{ 8 }$ Solve for $p$.
Answer: Multiply both sides by the left denominator. $\dfrac{ -6p + 5q }{ -{10} } = \dfrac{ 6p - 9r }{ 8 }$ $-{10} \cdot \dfrac{ -6p + 5q }{ -{10} } = -{10} \cdot \dfrac{ 6p - 9r }{ 8 }$ $-6p + 5q = -{10} \cdot \dfrac { 6p - 9r }{ 8 }$ Multiply both sides by the right denominator. $-6p + 5q = -10 \cdot \dfrac{ 6p - 9r }{ {8} }$ ${8} \cdot \left( -6p + 5q \right) = {8} \cdot -10 \cdot \dfrac{ 6p - 9r }{ {8} }$ ${8} \cdot \left( -6p + 5q \right) = -10 \cdot \left( 6p - 9r \right)$ Distribute both sides ${8} \cdot \left( -6p + 5q \right) = -{10} \cdot \left( 6p - 9r \right)$ $-{48}p + {40}q = -{60}p + {90}r$ Combine $p$ terms on the left. $-{48p} + 40q = -{60p} + 90r$ ${12p} + 40q = 90r$ Move the $q$ term to the right. $12p + {40q} = 90r$ $12p = 90r - {40q}$ Isolate $p$ by dividing both sides by its coefficient. ${12}p = 90r - 40q$ $p = \dfrac{ 90r - 40q }{ {12} }$ All of these terms are divisible by $2$ $p = \dfrac{ {45}r - {20}q }{ {6} }$